What is the third–degree polynomial function such that f(0) = –24 and whose zeros are 1, 2, and 3?

Question

anonymous · Answer

$$f(x)=(x-1)(x-2)(x-3)$$
is the simplest polynomial with the given roots. Find f(0), and adjust f(x) accordingly so that f(0) = -24.

anonymous · Answer

Since we know that zeroes are 1, 2, 3, then we know the equation looks something like this:

y = (x - 1)(x - 2)(x - 3) 

But the function must be -24 when x = 0, so let's see if our equation satisfies the condition when x = 0.

24 = (-1)(-2)(-3)
24 = -6 --> That's is wrong. Our answer should be 24, not -6. We now notice that our current function gives -6 for f(0). To turn that into 24, we can add an extra -4 at the front of the function so that when we get that -6 when x = 0, it gets multiplied by -4 to give us 24. Now let's see what the new function looks like and whether it works:

f(x) = -4(x - 1)(x - 2)(x - 3)
f(0) = -4(-1)(-2)(-3)
      = 24 --> Yes it works! 

Therefore, our third degree polynomial is: f(x) = -4(x - 1)(x - 2)(x - 3)